a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)

\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)

\(=\dfrac{3}{2}\left(\sqrt{5}-1\right)+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-2\sqrt{5}\)

\(=\dfrac{3}{2}\sqrt{5}-\dfrac{3}{2}-2\sqrt{5}+\dfrac{\sqrt{5}+1}{2}\)

\(=-\dfrac{1}{2}\sqrt{5}-\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\)

=-1

Bài 1: 

a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:

\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)

b) Ta có: \(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{x-9+\sqrt{x}+3-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+\sqrt{x}-6-x+2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\)

c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)

\(\Leftrightarrow3-\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}< 3\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)

ai giúp với ạ :<

2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{5\sqrt{x}-15}{3x-59}\)

1: Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}-\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\dfrac{25-x-x+9-x+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{-5\left(\sqrt{x}-3\right)}{-3x+59}\)

\(=\dfrac{5\sqrt{x}-15}{3x-59}\)​

2: Ta có: \(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

3: Ta có: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{x-1}\cdot\dfrac{1}{\sqrt{x}}\)

\(=\dfrac{2}{x-1}\)

\(E=\dfrac{\left|x-3\right|}{\left(x-3\right)\left(x+3\right)}\left(x+3\right)^2=\dfrac{\left|x-3\right|\left(x+3\right)}{x-3}\left(x\ne\pm3\right)\)

Với \(x>3\Leftrightarrow E=x+3\)

Với \(x< 3\Leftrightarrow E=-x-3\)

\(F=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\left(x\ge0;x\ne25\right)\\ F=\dfrac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)

\(A=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{5x\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\left(x>0;x\ne4\right)\\ A=\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{5x\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\\ A=\dfrac{10x\left(\sqrt{x}-2\right)}{\left(2\sqrt{x}+1\right)\left(2\sqrt{x}+3\right)}\)

\(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}}{2x-3\sqrt{x}-2}\right):\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)

\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\dfrac{2\sqrt{x}-4}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}\)

1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

Chỗ căn bị thiếu là √x nha.

Ta có: \(\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\dfrac{x-5\sqrt{x}-x+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}:\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{5}{\sqrt{x}+3}\)

\(A=2\left|2-\sqrt{5}\right|-\dfrac{8\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)

\(=2\left(\sqrt{5}-2\right)-\dfrac{8\left(3+\sqrt{5}\right)}{4}=2\sqrt{5}-4-2\left(3+\sqrt{5}\right)\)

\(=2\sqrt{5}-4-6-2\sqrt{5}=-10\)

\(B=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}\right)\)

\(=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)

\(=\dfrac{1}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{\sqrt{x}}\)